# Derivation of the Rayleigh Distribution Equation

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The Rayleigh Distribution makes the following assumptions:

• $$\bar{h} \sim \mathcal{N}(\bar{h},\sigma_h^2), \bar{v} \sim \mathcal{N}(\bar{v},\sigma_v^2)$$
• Horizontal and vertical dispersion are independent.
• $$\sigma_h = \sigma_v$$ (realistically $$\sigma_h \approx \sigma_v$$)
• $$\rho = 0$$
• No Fliers

for which the PDF is given by:
$$PDF_{r_i}(r) = \frac{r}{\sigma^2}e^{-r^2/2\sigma^2}$$

The PDF can be derived in the following two methods.

# Simplification of the Bivariate Normal Distribution

Using the assumptions above, the Rayleigh Distribution is easily simplified from the Bivariate Normal Distribution which has the equation:

$$f(h,v) = \frac{1}{2 \pi \sigma_h \sigma_v \sqrt{1-\rho^2}} \exp\left( -\frac{1}{2(1-\rho^2)}\left[ \frac{(h-\bar{h})^2}{\sigma_h^2} + \frac{(v-\bar{v})^2}{\sigma_v^2} - \frac{2\rho(h-\bar{h})(v-\bar{v})}{\sigma_h \sigma_v} \right] \right)$$

By substituting $$\rho = 0$$ the equation reduces to:

$$f(h,v) = \frac{1}{2 \pi \sigma_h \sigma_v } \exp\left( -\frac{1}{2}\left[ \frac{(h-\bar{h})^2}{\sigma_h^2} + \frac{(v-\bar{v})^2}{\sigma_v^2} \right] \right)$$

Since $$\sigma_h$$ and $$\sigma_v$$ are equal, substitute $$\sigma$$ for each, then collect terms in the exponential, after which the equation reduces to:

$$f(h,v) = \frac{1}{2 \pi \sigma^2 } \exp\left( -\left[ \frac{(h-\bar{h})^2 + (v-\bar{v})^2}{2\sigma^2} \right] \right)$$

Letting $$r^2 = (h-\bar{h})^2 + (v-\bar{v})^2$$ the equation becomes:

$$f(h,v) = \frac{1}{2 \pi \sigma^2 } \exp\left( - \frac{r^2}{2\sigma^2} \right)$$

By transforming to the polar coordinate system one has:

$$f(h,v) = \frac{1}{2 \pi \sigma^2 } \exp\left( - \frac{r^2}{2\sigma^2} \right)$$